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Re: Endorsement of Wolfgang Mueckenheim from a serious mathematician
Posted:
Jan 31, 2013 4:39 AM
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On 31 Jan., 10:31, Virgil <vir...@ligriv.com> wrote:
> In article
> <65959390-0681-4d8a-8e3c-7b1e8dad9...@w7g2000yqo.googlegroups.com>,
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> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 31 Jan., 01:58, Virgil <vir...@ligriv.com> wrote:
> > > In article
> > > <397a90fb-2e3e-411f-ae40-2365cadd1...@b11g2000yqh.googlegroups.com>,
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> > > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > I have developed a new proof-technique, namely proof by ignorance
>
> > > That is hardly new for Wolkenmuekenheim, but has been the ONLY standard
> > > there for years.
>
> > Thinks are easy if things are easy.
> > Consider a box with a dozen different pralines. If I take three out of
> > the boxh and return it to you, you can decide which I did not take out
> > by looking at the remaining ones.
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> > Consider a Binary Tree. When I remove or colour some paths, you can
> > decide which are remaining by looking at the uncoloured paths.
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> > And finally consider a Binary Tree with uncountably many paths. When I
> > colour a countable set of paths, then you have to decide by looking at
> > the remainings which paths have survived.
>
> While WM may include every node in his countable set of paths, he does
> not, and cannot, simultanteously include every path.
I include every possible combination of nodes that reside at finite
levels. If there are not nodes residing at infinite levels, then I
include every path. If you don't agree, show another path.
But remember, I do not stop at a node at a finite level, but in
addition, I append every possible tail, i.e., every tail that can be
defined by a finite word.
I beg every amount of money that you will not be able to find a
missing path. Hence yor assertion is based only on (counterfactual)
belief. A typical matheologian.
Regards, WM
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